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final_solution

# final_solution - ECE 147A FEEDBACK CONTROL SYSTEMS THEORY...

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ECE 147A FEEDBACK CONTROL SYSTEMS - THEORY AND DESIGN F09 Final Exam Closed book and notes. No calculators or cell phones. Show all work. Name: Solution Problem 1: /15 Problem 2: /15 Problem 3: /15 Problem 4: /15 Problem 5: /15 Problem 6: /15 Total: /90 Miscellaneous: tan 1 (1 / 3) = 18 . 435 degrees, tan 1 (1 / 6 . 1623) = 9 . 2175 degrees, tan 1 (1) = 45 degrees. tan 1 (2) = 63 . 435 degrees, tan 1 (3) = 71 . 565 degrees, tan 1 (5) = 78 . 69 degrees, sin(45 ) = 2 / 2 0 . 7, sin(50 ) 0 . 77, sin(55 ) 0 . 82, sin(60 ) = 3 / 2 0 . 87. A lead compensator λ ( s + z ) s + λz ( λ > 1) introduces a maximum possible positive phase φ max satisfying λ = 1+sin( φ max ) 1 sin( φ max ) at the frequency λ · z where the magnitude is λ . L (cos( ωt )) = s s 2 + ω 2 , L ( t sin( ωt )) = 2 ωs ( s 2 + ω 2 ) 2

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Name: Problem 1 1. How many roots of the equation s 5 + 5 s 4 + 10 s 3 + 8 s 2 + 13 s + 5 = 0 have positive real part? We construct the Routh array for the given polynomial: s 5 1 10 13 s 4 5 8 5 s 3 5 / 6 7 10 0 s 2 7 6 35 0 s 1 6 -185 0 0 s 0 35 0 0 2 sign changes in the 1 st column indicate that the equation has 2 roots with positive real part. 2. A Nyquist plot for the transfer function L ( s ), which has three unstable poles, is shown below. For each value of K (both positive and negative), indicate the number of roots with positive real part for the equation 1 + KL ( s ) = 0. -0.25 -0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 0.2 0.25 -0.25 -0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 0.2 0.25 Nyquist Diagram Real Axis Imaginary Axis
Name: (Formerly blank page) The number of roots with positive real part is equal to P + N where P = 3 is the number of unstable poles of L ( s ) and N is the number of encirclements of the ( 1 /K, 0) point in the complex plane (counterclockwise encirclements corresponding to N < 0 ). We see that the point ( 1 /K, 0) is: 1. encircled zero times when 1 /K < 0 . 2 and 1 /K > 0 . 15 ; therefore, the equation has 3 roots with positive real part for K ( 20 / 3 , 5) . 2. encircled one time in the counterclockwise direction when 0 . 2 < 1 /K < 0 . 06 ; therefore, the equation has 2 roots with positive real part for K (5 , 50 / 3) . 3. encircled three times in the counterclockwise direction when 0 . 06 < 1 /K < 0 ; therefore the equation has 0 roots with positive real part for K > 50 / 3 . 4. encircled two times in the counterclockwise direction when 0 < 1 /K < 0 . 15 ; therefore, the equation has 1 root with positive real part for K < 20 / 3 .

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Name: Problem 2 Consider the nonlinear system with parameter ω negationslash = 0 given as ˙ x 1 = x 2 + cos( ω 2 x 1 ) ˙ x 2 = sin( ω 2 x 1 ) + u y = x 1 . 1. Determine all valid operating points corresponding to an output value of zero. For the output to be zero, we must have x 1 = 0 . Since cos(0) = 1 , this implies that x 2 = 1 . Since sin(0) = 0 , this implies that u = 0 . So, there is only one valid operating point. It is given as (0 , 1 , 0) . 2. Find the transfer function corresponding to one of the operating points. Express your answer in terms of ω . Hint: use the fact that bracketleftBigg a b c d bracketrightBigg 1 = bracketleftBigg d b c a bracketrightBigg ad bc .
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